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3 years ago

Slope intercept form of 7x-2y=-8

Mathematics

2 answers:

Vlada [557]3 years ago

5 0

7x-2y=-8

Slope= 7/2

x-intercept= -1.143

y-intercept= 4

eimsori [14]3 years ago

3 0

The correct way to do this would be like this.

step one make it look like this -2y=-7x - 8 the 7 becomes -7 when you move it to the other side

step two divide by -2 to get y= 7/2x + 4

So now you get the answer witch is y= 7/2x + 4

Please mark me as Brainiest if this helped

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Holly and Tamar completed the work in the table to determine if a triangle with side lengths of 40, 42, and 58 is a right triang

PIT_PIT [208]

Simple Pythagorean theorem, And if you double check Tamara's work she is correct. Therefor B is the correct answer.

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2 years ago

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All four B points are equidistant from points A and C. Are they all midpoints of AC? Justify your answer.

Rina8888 [55]

Answer:

Step-by-step explanation:

For a point to be the midpoint of a line segment, it must bisect it into two equal segments and be on the line segment (hence, colinear with the endpoints). All four B points are equidistant from points A and C, but aren’t colinear with A and C. Therefore, they aren’t all midpoints of line segment AC.

I hope this helps! :)

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2 years ago

Whitch ordered pairs are solutions to the inequality y-4x≤-6

konstantin123 [22]

Answer:

y </ 4x - 6

Step-by-step explanation:

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3 years ago

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How do you know that 21 over 30 is greater than 2 over 3

Bond [772]

The problem deals with fractions comparison, lets do it:
21/30 > 2/3
we begin solving:
21 > (2/3)*30
21 > 2*10
21 > 20
therefore the proposed inequality is true, 21/30 > 2/3

You can solve as well getting same denominator for both fractions and comparing directly, in this case we need to get 2/3 to be divided by 30:
2/3 = (10/10)(2/3) = 20/30
So we have:
21/30 > 2/3
which is equal to:
21/30 > 20/30
and we compare directly because both fractions are divided by the same number, and we can see that the inequality is true.

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3 years ago

Determine the formula for the nth term of the sequence: -2,1,7,25,79,...

rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

$x_n = a n^4 + b n^3 + c n^2 + d n + e$

From the given known values of the sequence, we have

$\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}$

Solving the system yields coefficients

$a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4$

so that the n-th term in the sequence might be

$\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}$

Then the next few terms in the sequence could very well be

$\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}$

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of $\{x_n\}$ eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of $\{x_n\}$ by $\Delta^{k}\{x_n\}$ . Then

• 1st-order differences:

$\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}$

• 2nd-order differences:

$\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}$

• 3rd-order differences:

$\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}$

• 4th-order differences:

$\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}$

From here I made the assumption that $\Delta^4\{x_n\}$ is the constant sequence {15, 15, 15, …}. This implies $\Delta^3\{x_n\}$ forms an arithmetic/linear sequence, which implies $\Delta^2\{x_n\}$ forms a quadratic sequence, and so on up $\{x_n\}$ forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

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2 years ago